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Projectile Motion- magnitude of final velocity

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a cannon on a 50m high cliff. The cannon shoots a cannonball with an initial speed of 200m/s at an angle of 30°, causing the cannonball to land 41m away from the cannon. What is the magnitude of the velocity of the cannonball the instance just before it hits the ground?
    a) -205m/s
    b) 205m/s
    c) 173m/s
    d) 267m/s

    2. Relevant equations
    ax=0 ay=-g
    1) Xf=Xi+Vixt
    2) Yf=Yi+Viyt-½gt2
    3) Vfy=Viy-gt
    4) Vfy2=Viy2-2g(Yf-Yi)

    3. The attempt at a solution
    Vi=200m/s
    Vix=200cos30 Viy=200sin30
    Xi=0m Yi= 50m
    Xf=41m Yf= 0m

    In order to find the magnitude of Vf, I need to find its components, Vfx and Vfy. In projectile motion, Vfx=Vix=200cos30, so I just need to find Vfy.

    Using equation 1 to find time: t=41/(200cos30)= 0.2367s
    I then plugged time into equation 3 to find Vfy=200sin30-(9.8)(0.2367)= 97.68m/s

    So Vfx=200cos30= 173.2m/s and Vfy= 97.68m/s so
    Vf= √(Vfx2+Vfy2)= √(173.22+97.682)= 199 m/s

    I don't see where I'm going wrong, but 199m/s is not one of the 4 multiple choice answers. I would greatly appreciate any help.
     
  2. jcsd
  3. Apr 3, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hi, Nelson2436. Welcome to PF!

    The 41 m given in the problem just doesn't seem right. You should actually be able to find where the cannon ball lands from the other information given in the problem. Did you state the problem exactly as given?
     
  4. Apr 3, 2015 #3
    Yes, I stated the problem exactly as given. This problem is driving me crazy.
     
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